Question

Помогите пожалуйста решить.

Answer 1

Ответ:

Объяснение:

2)

$10t^2 + 3t - 1 = 0\\D = 9 - 4 * 10 * (-1) = 9 + 40 = 49\\\sqrt{D} = 7\\ t_{1} = \frac{-3 + 7}{2 * 10} = \frac{4}{20} = \frac{1}{5} \\\\ t_{2} = \frac{-3 - 7}{2 * 10} = \frac{-10}{20} = -\frac{1}{2}\\\left \{ {{cosx = \frac{1}{5}} \atop {cosx = -\frac{1}{2}\\}} \right.\\\\cosx = \frac{1}{5}\\ x = +- arccos(\frac{1}{5}) + 2\pi k\\cosx = -\frac{1}{2} \\x = +- arccos(-\frac{1}{2} ) + 2\pi k\\x = \pi - arccos(\frac{1}{2}) + 2\pi k\\x = \pi - \frac{\pi }{3} + 2\pi k\\x = \frac{2\pi }{3} + 2\pi k$

3)

$5sinx + 2cosx = 0 | : cosx\\\\5tgx + 2 = 0\\5tgx = -2 | : 5\\tgx = -\frac{2}{5} \\x = arctg(-\frac{2}{5}) + 2\pi k\\x = \pi - arctg(\frac{2}{5}) + 2\pi k$